In how many ways can 5 boys and 5 girls be ar | vedclass

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(C) First,arrange the $5$ boys in a row,which can be done in $5!$ ways.There are $6$ possible gaps created between and at the ends of the boys (repres

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$5! 	imes 6!$

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(C) First,arrange the $5$ boys in a row,which can be done in $5!$ ways.There are $6$ possible gaps created between and at the ends of the boys (represented as _ $B_1$ _ $B_2$ _ $B_3$ _ $B_4$ _ $B_5$ _).To ensure no two girls are together,we must place the $5$ girls in these $6$ gaps.The number of ways to choose and arrange $5$ girls in $6$ gaps is given by $^6P_5$.Therefore,the total number of arrangements is $5! 	imes ^6P_5 = 5! 	imes \frac{6!}{(6-5)!} = 5! 	imes 6!$.

In how many ways can $5$ boys and $5$ girls be arranged in a row such that no two girls are together?

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